HWKey (18) - how the electrons move in these 2 propagation...

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Problem of the Day #17 – Answer Key Deadline : 3:00 p.m., Monday, 7/7/08 LATE WORK WILL NOT BE ACCEPTED OR GRADED!!! This problem is worth a total of 20 raw points. Alkyl halides can be converted into alkanes via reaction with tributyltin hydride, (C 4 H 9 ) 3 Sn–H, in the presence of light (h ν ). The reaction of cyclohexyl chloride to give cyclohexane is shown below as a representative example. This reaction occurs via a radical chain reaction mechanism. Cl + (C 4 H 9 ) 3 Sn H h ν H + (C 4 H 9 ) 3 Sn Cl (a) Use “fishhook” arrows to propose an initiation step for this reaction. Hint : Light (h ν ) induces homolytic cleavage of a weak bond. The Sn-H bond is rather weak, and is therefore highly susceptible to hemolytic cleavage. (b) The propagation part of the reaction mechanism involves two steps. Use “fishhook” arrows to show
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Unformatted text preview: how the electrons move in these 2 propagation steps. Hint : Both of the products shown above are formed in the propagation cycle. The radical consumed in the first propagation step is regenerated in the second propagation step. (c) There are three potential termination steps in this mechanism. Write two of them, again being sure to use fishhook arrows to show how the electrons move. Remember, a valid termination step involves the combination of any two radicals involved in the propagation part of the reaction mechanism. (C 4 H 9 ) 3 Sn H (C 4 H 9 ) 3 Sn + H + (C 4 H 9 ) 3 Sn + Cl (C 4 H 9 ) 3 Sn Cl (C 4 H 9 ) 3 Sn H + H + (C 4 H 9 ) 3 Sn (C 4 H 9 ) 3 Sn + Sn(C 4 H 9 ) 3 (C 4 H 9 ) 3 Sn Sn(C 4 H 9 ) 3 + (C 4 H 9 ) 3 Sn + (C 4 H 9 ) 3 Sn...
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